An ionic compound AB has a fluorite-type structure. If the radius of B is 200 pm, what is the ideal radius of A? Options: A: 82.8 pm B: 146.4 pm C: 40 pm D: 45 pm

The correct answer is: D

The solution can be understood as follows:

In an ideal fluorite structure, the cation ($A$) is present in a tetrahedral void. Hence, we can use the relationship for the ratio of ionic radii in this structure:

$$\frac{r_c}{r_a} = 0.225$$

Given that the radius of $B$ ($r_c$) is $200$ pm, we can solve for the radius of $A$ ($r_a$):

$$r_a = \frac{r_c}{0.225} = \frac{200 \text{ pm}}{0.225} \approx 44.4 \text{ pm}$$

Since the closest option to this calculation is 45 pm, the ideal radius of $A$ is:

Option D: 45 pm